Gravitational time dilation in weak field

Question

A plane flies at a constant height $h$. I should calculate, which speed $v$ the plane shoud have, such that an observer on ground sees the a clock whithin the plane tick in the same rate as a clock on ground.

In the Newtonian limit, I have the following relations:

$$\mathrm{d}\tau_{\mathrm{ground}}=\sqrt{1+\frac{2\Phi_{\mathrm{ground}}}{c^{2}}}\mathrm{d}t_{\mathrm{ground}}=\sqrt{1-\frac{2GM}{c^{2}R_{\mathrm{Earth}}}}\mathrm{d}t_{\mathrm{ground}}$$

$$\mathrm{d}\tau_{\mathrm{plane}}=\sqrt{1+\frac{2\Phi_{\mathrm{plane}}}{c^{2}}-\frac{v^{2}}{c^{2}}}\mathrm{d}t_{\mathrm{plane}}=\sqrt{1-\frac{2GM}{c^{2}(R_{\mathrm{Earth}}+h)}-\frac{v^{2}}{c^{2}}}\mathrm{d}t_{\mathrm{plane}}$$

In this formulas, $t$ is the coordinate time (e.g. the proper time of an observer at infinity) and $\tau$ the proper time of the observers.......

How can I determine the velovity?

EDIT: Is it right when I say that $\mathrm{d}t_{\mathrm{plane}}=\mathrm{d}t_{\mathrm{ground}}$, because the interval is the same for the coordinate time (observer at infinity (Phi=0))....Because then I find $$\frac{\mathrm{d}\tau_{\mathrm{plane}}}{\mathrm{d}\tau_{\mathrm{ground}}}\approx 1+\frac{\Delta\Phi}{c^{2}}-\frac{v^{2}}{2c^{2}} =^{!}1$$ and therefore: $v\approx \sqrt{2\Delta\Phi}=\sqrt{2GM(\frac{1}{R}-\frac{1}{R+h})}$

Answer 1

You need to solve for $v$:

$$\frac{g_{t t}(r_o)}{g_{t t}(r_p)}=\gamma^2 \ \to \ \frac{1-r_s/r_o}{1-r_s/r_p}=1-\frac{v^2}{c^2}$$

where $r$ is the radius from the center of mass, $r_o$ the radius where the observer is, $r_p=r_o+h$ the radius where the plane is at, $r_s$ the Schwarzschild radius and $v$ the relative velocity between the observer and the plane. Then you get

$$v=c \sqrt{\frac{r_s (r_p-r_o)}{r_o (r_p-r_s)}}$$

Here we assume that the velocity of the plane is perpendicular to the observer, so that we can neglect the radial Doppler shift. In the slow velocity and large radius limit that fits your result.